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hence \\^y\\ ^ \\ng(x)\\  \\7ig(y}\\, and therefore

^ \\ng(x)\\ - \\ng(y)\\.

Since A/ng is dense in H^ 9 we infer that there exists a vector   x e H5 such
that |/^y({0})| = ( ; x 1 7r/y)) for all y e A (6.3.2), and we have


This shows first that E  x depends only on ng(x) and can therefore be written
in the form W0  ng(x), where W0 is a linear mapping of A/ng into H^; also, we
have || WQ - ng(x)\\ g |]rcff(*)ll, so that W0 is continuous, and extends to a con-
tinuous operator W e 5f(H^). Next, replacing x by zx in (1 7) and noting
that V-ng(zx) = (VUg(z)) ' xg(x) for Fe ja^, we obtain

Since the function .%'*-+ (9 Ug(z'))(a>~1(x'y) is zero at the point 0, we see that
/*z*,/{0}) = 0 fr a^ z  A> or equivalently (W - ng(zx) \ ng(y)) = 0. Since the
elements ng(y) are dense in H^, it follows that W - ng(zx) = 0 for all z and all
x in A; but by virtue of the condition (N) (which is used only at this point of
the proof) the ng(zx) form a total set in H^ . Hence we have W = 0, and the
proof of (i) is complete.

(15.9.120)   Proof of (ui).

To show that the support of mg is the whole of S^ , it is enough to prove
that, for each x e S^ and each open neighborhood V of x in S^, there exists at
least one continuous function F ^ 0, with support contained in V, and such
that mg(F) ^ 0. Suppose that this is not the case. Then for every F as above
we have

f Fft) dmxi,(x) = . [

= 0

for all x, y in A. But by virtue of the Gelfand-Neumark theorem (1 5.4.1 4), F is
the restriction to S^ of a function of the form ^Ko to"1, where Ve^'g, and
by ( we have (V- ng(x) \ ng(y)) = 0 for all x, y in A. Since Afng is
dense in Hff , it follows that V = 0 and therefore F = 0, which is absurd (4.5.2).

(   Proof of (IN).

By virtue of (iii), the vector space ^c(S^) n &l(Sg , mg) (and therefore also
its subspace tfc(Sg)) c&n be identified (algebraically) with a dense subspace
of I|(S,,/,) (13.11.6). Since on the other hand the relation ( can
be put in the form (ng(x) \ng(yj) = *(x)Jft) dmg(x) and therefore definesions xnxn therefore converge uniformly on S^ to G, and hence,